Question

A bullet fired into a target loses half its velocity after penetrating $25cm$. How much further will it penetrate before coming to rest?

• A. $35cm$
• B. $25cm$
• C. $8.3cm$
• D. $75cm$

Answer 1

The correct option is C

$8.3cm$

Step 1: Given data

Let us assume the initial velocity is $v$

The final velocity after penetrating is$\frac{v}{2}$

Distance travelled by the bullet is 25 cm.

Step 2: Finding the distance travelled by bullet

In the first case,

$$\begin{gathered} Changeinkineticenergy=workdone\times displacement \\ \frac{1}{2}m{v}^2-\frac{1}{2}m{\left(\frac{v}{2}\right)}^2=W\times 25 \\ \frac{1\times 4}{2\times 4}m{v}^2-\frac{1}{8}m{v}^2=W\times 25 \\ \frac{3}{8}m{v}^2=W\times 25\left(i\right) \end{gathered}$$

In the second case,

$$\begin{gathered} Changeinkineticenergy=workdone\times displacement \\ \frac{1}{2}m{\left(\frac{v}{2}\right)}^2-\frac{1}{2}m{\left(0\right)}^2=W\times S \\ \frac{1}{8}m{v}^2=W\times S\left(ii\right) \end{gathered}$$

Step 3: Solving the two equations

Dividing equation (ii) by equation (i), we get

$$\begin{gathered} \frac{W\times S}{W\times 25}=\frac{{\displaystyle \frac{1}{8}m{v}^2}}{{\displaystyle \frac{3}{8}m{v}^2}} \\ \frac{S}{25}=\frac{1}{3} \\ S=\frac{25}{3}=8.3cm \end{gathered}$$

The distance travelled by bullet before coming to rest is $8.3cm$

Hence, option (c) is correct.